Question

A force of magnitude 5 unit acting along the vector $2\hat{i}-2\hat{j}+\hat{k}$ displaces the point of application from (1, 2, 3) to (5, 3, 7). Then the work done is:

• A. 50/7 unit
• B. 50/3 unit
• C. 25/3 unit
• D. 25/4 unit
• E. 3/50 unit

Answer 1

Answer: (B)

$\therefore$ $\overrightarrow{F}=\frac{5(2\hat{i}-2\hat{j}+\hat{k})}{\sqrt{4+4+1}}=\frac{5}{3}(2\hat{i}-2\hat{j}+\hat{k})$ and $\overrightarrow{d}=(5\hat{i}+3\hat{j}+7\hat{k})-(\hat{i}+2j+3\hat{k})$ $=4\hat{i}+\hat{j}+4\hat{k}$ $\therefore$ $W=\overrightarrow{F}.\overrightarrow{d}=\frac{5}{3}(2\hat{i}-2\hat{j}+\hat{k})$ $(4\hat{i}+\hat{j}+4\hat{k})$ $=\frac{5}{3}(8-2+4)=\frac{50}{3}unit$